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Dual vector space - Lagrange Interpolating Polynomial

  1. Feb 21, 2013 #1
    I think I solved it a week ago, but I didn't write down all the things and I want to be sure of doing the things right, plus the excersise of writing it here in latex helps me a loot (I wrote about 3 threads and didn't submited it because writing it here clarified me enough to find the answer for my self xD). So feel free to correct me.

    Lets be [itex]\alpha _{0},...,\alpha _{n} \in K, \alpha _{i} \neq \alpha _{j}[/itex] if [itex] i\neq j[/itex]

    To every [itex]i[/itex], [itex]0 \leq i \leq n[/itex], we define [itex]\varepsilon _{\alpha _i}:K_{n}[X]\rightarrow K [/itex] as [itex]\varepsilon _{\alpha _i}(P)=P(\alpha _i)[/itex].

    i) Prove that [itex]B = \left \{ \epsilon _{\alpha _{0}},...,\epsilon _{\alpha _{n}} \right \}[/itex] is a basis of [itex](K_{n}[X])^{*}[/itex]

    ii) Lets be [itex]B = \left \{ P_{0},...,P_{n} \right \}[/itex] the basis of [itex]K_{n}[X][/itex] such that [itex]B^{*}=B_{1}[/itex]. Prove that the polynomial

    [itex]P = \sum _{i=0}^{n} \beta _i . P_i[/itex]

    is the only polynomial in K[X] of grade minor or equal to n such that,

    [itex]\forall i, 0 \leq i \leq n, P(\alpha _{i}) = \beta _{i}[/itex]

    This polynomial is named the Lagrange's Interpolating Polynomial (irrelevant :P).

    iii) Prove that there're real numbers [itex]\left \a_{0},...,a_{n}{ \right \}[/itex] such that, to all [itex]P \in \mathbb{R}_{n}[X][/itex]
    [itex]\int _{0}^{1}P(x)dx=\sum _{i=0}^{n}a_{i}.P(\alpha _{i})[/itex]
    Find [itex]a_{0}, a_{1}[/itex] and [itex]a_{2}[/itex] when [itex]n = 2, a_{0}=1, a_{1}=\frac{1}{2}[/itex] and [itex] a_{2}=0[/itex]

    Relevant proposition:
    Lets be V a K-vector space of dimensión n, and lets be V* it's dual space.
    Lets be [itex]B_{1}=\left \{ \varphi _{1},...\varphi _{n} \right \}[/itex] a basis of V*. Then there's only one basis [itex]B = \left \{ v _{1},...v _{n} \right \}[/itex] of V such that:
    [itex]B^{*}=B_{1}[/itex]

    i) Suppose that:
    [itex]\sum _{i=0}^{n}\alpha _{i}\varepsilon _{i}=0[/itex]
    Then, to all [itex]P \in K_n[X][/itex], we have: [itex]\sum _{i=0}^{n}\alpha _{i}\varepsilon _{i}(P)=\sum _{i=0}^{n}P(\alpha _{i})=0[/itex]
    To every [itex]\forall i, 0 \leq i \leq n[/itex]
    We take the polynomial:
    [itex]\prod _{i=0,{i\neq }j}^{n}(X-\alpha _{i})[/itex]
    And evaluating it we know [itex]\alpha _{j}=0[/itex], because [itex]P(\alpha _{i})=0[/itex] to every i≠j [itex](P(\alpha _{j})[/itex] can't be zero because [itex]\alpha _{j} \neq \alpha _{i}[/itex] if i≠j
    Then, we have the linear independency of the set, and as n+1 elements, then it's generated space has dimention n+1 and it's a subset of [itex](K_{n}[X])*[/itex], so it's a basis of [itex](K_{n}[X])*[/itex] (because it also has dimension n+1).
    ii) So we know [itex]B_{1}[/itex] is a basis of [itex]K_n[X][/itex], then for the proposition we know that there's only one basis
    [itex]B= \left \{ P_{0},...,P_{n} \right \}[/itex] such that [itex]B^{*}=B_{1}[/itex]
    So, by the definition of dual basis, [itex]\epsilon _{\alpha _{i}}(P_{j})=P_{j}(\epsilon _{\alpha _{0}})=[/itex]
    1 if i=j
    0 if i≠j
    Then [itex]P(\epsilon _{\alpha _{j}})=\sum _{i=0}^{n} \beta _i . P_i(\epsilon _{\alpha _{j}})=\beta _{i}[/itex]

    iii)
    I think I'm stuck in 3, but I'll think later after sleeping, I'm probably failing due to being so sleepy.
    [itex][/itex]
    [itex][/itex]
    [itex][/itex]
    [itex][/itex]
    [itex][/itex]
     
    Last edited: Feb 21, 2013
  2. jcsd
  3. Feb 22, 2013 #2
    I didn't find the edit button in my first post, I assume someone blocked it so I bump the thread when I finish it instead of buring it in the bottom of the forum's abyss.
    If [itex]P=\sum _{i=0}^{n}\beta _{i} P_{i} [/itex] then
    [itex]\int _{0}^{1}\sum _{i=0}^{n} \beta _{i} . P_{i} (x)dx = \sum _{i=0}^{n} \beta _{i}.\int _{0}^{1}P_{i} (x)dx[/itex]
    [itex]\sum _{i=0}^{n}a_i.P(\alpha _{i})=\sum _{i=0}^{n}a_i.\sum _{j=0}^{n} \beta _{i} P_{i} (\alpha _{i})=[/itex]
    [itex]\sum _{i=0}^{n}a_{i}.\beta _{i}[/itex]
    Let's be [itex]a_{i}=\int _{0}^{1}P_{i} (x)dx[/itex] to all [itex]0 \leq i \leq n[/itex]
    Then
    [itex]\sum _{i=0}^{n} \frac{\beta _{i}}{i+1}=\sum _{i=0}^{n}a_{i}.\beta _{i}[/itex]


    Let's do find [itex]a_{0}, a_{1}[/itex] and [itex]a_{2}[/itex] when [itex]n = 2, a_{0}=1, a_{1}=\frac{1}{2}[/itex] and [itex] a_{2}=0[/itex]
    [itex]\left\{\begin{matrix}
    P_{0}(\alpha _{0})=1\\
    P_{0}(\alpha _{1})=0\\
    P_{0}(\alpha _{2})=0
    \end{matrix}\right.[/itex]
    [itex]\left\{\begin{matrix}
    P_{0}(1)=1\\
    P_{0}(\frac{1}{2})=0\\
    P_{0}(0)=0
    \end{matrix}\right.[/itex]
    As we know two roots of [itex]P_{0}[/itex] we only need to multiply it for a constant to make it equal to 1 when evaluated in 1.
    [itex]-\frac{(X-\frac{1}{2})(X)}{2}[/itex]

    [itex]
    \left\{\begin{matrix}
    P_{1}(1)=0\\
    P_{1}(\frac{1}{2})=1\\
    P_{1}(0)=0
    \end{matrix}\right.[/itex]
    [itex]-4(X-1)(X)[/itex]

    [itex]\left\{\begin{matrix}
    P_{2}(1)=0\\
    P_{2}(\frac{1}{2})=0\\
    P_{2}(0)=1
    \end{matrix}\right.[/itex]

    [itex]2(X-\frac{1}{2})(X-1)[/itex]
    [itex][/itex]
    So we only need to evaluate the integrals of these polynomials.
    Then:
    [itex]a_{0}=-\frac{1}{24}[/itex]
    [itex]a_{1}=\frac{2}{3}[/itex]
    [itex]a_{2}=\frac{1}{6}[/itex]

    PD: The excersise is from http://mate.dm.uba.ar/~jeronimo/algebra_lineal/AlgebraLineal.pdf page 115 Ejercise 16, this isn't the book used in my Linear Algebra's course in the dual vector space subject but I find the ejercises were pretty interesting to do.
     
    Last edited: Feb 22, 2013
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